How Can I Find the Taylor Series Around a Specific Point Using Trigonometric Identities and Integral Calculus?

When working with integrals involving trigonometric functions, it is often useful to simplify the integrand using standard trigonometric identities before applying the Taylor series expansion. Let's walk through an example involving the integral of a function with a cos(2t^2) term. This process involves several steps: simplifying the integrand, using known expansions, and applying the fundamental theorem of calculus.

Simplifying the Integrand Using Trigonometric Identities

The given integral involves cos(2t^2), which can be expressed using the standard identity for the cosine of a sum:

cos(A B) cos(A)cos(B) - sin(A)sin(B)

Applying this identity to cos(2t^2), we have:

cos(2t^2) cos(1)cos(t^2) - sin(1)sin(t^2)

Let's substitute this into the integrand:

cos(1)cos(t^2) - sin(1)sin(t^2)

To integrate this, we can break it down further:

cos(1)sin(t^2)cos(t^2) - sin(1)sin(t^2)sin(t^2)

This can be simplified using the double-angle identities:

sin(2t^2) 2sin(t^2)cos(t^2)

Thus, the expression becomes:

1/2 * cos(1)[sin(3t^2) - sin(t^2)] - 1/2 * sin(1)[cos(t^2) - cos(3t^2)]

Applying Taylor Series Expansions

Now that we have simplified the integrand, it is time to apply the Taylor series expansions for sine and cosine functions. The general forms for the Taylor series expansions are:

sin(u) u - u^3/3! u^5/5! - ...

cos(u) 1 - u^2/2! u^4/4! - ...

Applying these to our simplified expression, we start with:

sin(3t^2) - sin(t^2) 3t^2 - (3t^2)^3/3! - (3t^2)^5/5! ... - t^2 (t^2)^3/3! (t^2)^5/5! ...

And:

cos(t^2) - cos(3t^2) 1 - (t^2)^2/2! - (t^2)^4/4! - (t^2)^6/6! ... - 1 (3t^2)^2/2! (3t^2)^4/4! (3t^2)^6/6! ...

Simplifying these, we get a more specific expansion:

sin(3t^2) - sin(t^2) 3t^2 - 27t^6/6 - 243t^{10}/120 ... - t^2 t^6/6 t^{10}/120 ...

cos(t^2) - cos(3t^2) 1 - t^4/2 - t^8/24 - t^{12}/720 ... - 1 9t^4/2 - 81t^8/24 729t^{12}/720 ...

This provides a good starting point for the computation. To fully expand this and find the Taylor series, you would need to continue the series by including more terms as necessary.

Conclusion and Next Steps

With these expansions, you can now proceed to compute the Taylor series for the given integral. The fundamental theorem of calculus tells us that the integral of our function can be found using the derivative of the integrand, without actually integrating. In practice, this means that we can use the above expansions to find the Taylor series of the integral function.

By substituting the expansions into the integrand and performing the necessary algebraic operations, you can derive the Taylor series of the integral around the specific point of interest. This process is often useful in approximation and solving complex integrals that are difficult to integrate directly.

Keywords: Taylor series, integral, trigonometric identities, standard expansions, calculus